Answer
$s_{fg}=1.053\text{ Btu/lbmâ‹…R}$
Work Step by Step
From the Clapeyron equation,
$$
\left(\frac{d P}{d T}\right)_{s a t}=\frac{h_{f g}}{T v_{f g}}=\frac{s_{f g}}{v_{f g}}
$$ Solving for $s_{f g}$, $$
s_{f g}=\frac{h_{f g}}{T}=\frac{(250 \mathrm{Btu}) /(0.5 \mathrm{lbm})}{475 \mathrm{R}}=\mathbf{1 . 0 5 3 B t u} / \mathrm{lbm} \cdot \mathbf{R}
$$ Alternatively, $$
s_{f g}=\left(\frac{d P}{d T}\right)_{\mathrm{sat}} v_{f g}=(1.896 \mathrm{psia} / \mathrm{R}) \frac{1.5 \mathrm{ft}^3}{0.5 \mathrm{lbm}}\left(\frac{1 \mathrm{Btu}}{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}\right)=1.053\ \mathrm{Btu} / \mathrm{lbm} \cdot
R$$
